What is the slope of the line tangent to $f(x) = -2x^{2}-4x+3$ at $x = 1$ ?
Solution: The slope of the tangent line is $ \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{(-2(x+\Delta x)^{2}-4(x+\Delta x)+3) - (-2x^{2}-4x+3)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{(-2(x^{2}+2x \Delta x+\Delta x^{2})-4(x+\Delta x)+3) - (-2x^{2}-4x+3)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{-2x^{2}-4(x \Delta x)-2\Delta x^{2}-4x-4(\Delta x)+3+2x^{2}+4x-3}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{-4(x \Delta x)-2\Delta x^{2}-4(\Delta x)}{\Delta x}$ $ = \lim_{\Delta x \to 0} -4x-2(\Delta x)-4$ $ = -4x-4$ $ = (-4)(1)-4$ $ = -8$